 Solution

Let $\zeta$ be a primitive fifth root of unity. Over ${\mathbb{F}}_p$, the Frobenius automorphism acts by $\zeta \mapsto {\zeta }^p$, so the splitting field of $X^5-1$ has degree equal to the multiplicative order of $p$ modulo $5$. Its Galois group is generated by Frobenius and is cyclic. The possible orders are therefore $1$ for $p\equiv 1$, $4$ for $p\equiv 2,3$, and $2$ for $p\equiv 4(\mathrm{mod}5)$.

Put $\alpha =5^{1/5}$. The polynomial $X^5-5$ satisfies the Eisenstein criterion at $5$, so $[\mathbb{Q}(\alpha ):\mathbb{Q}]=5$. Also $[\mathbb{Q}(\zeta ):\mathbb{Q}]=\phi (5)=4$. The intersection of these two fields has degree dividing both $5$ and $4$, hence is $\mathbb{Q}$. Since the splitting field is $\mathbb{Q}(\alpha ,\zeta )$,

Label the five roots by $\alpha {\zeta }^i$, $i\in {\mathbb{F}}_5$. The Galois group acts through the affine group over a finite field:Its possible nontrivial cycle types are $5$, $1+2+2$, and $1+4$; in particular, no element has a three-cycle. By the Dedekind factorization cycle test, the reduction of $X^5-5$ modulo any prime $p\ne 5$ therefore has no irreducible cubic factor.

The multiplier of the Frobenius element is $a=p(\mathrm{mod}5)$. If $p\equiv 2$ or $3$, then $a$ has order four; every affine map with this multiplier has one fixed point and one four-cycle. Hence $X^5-5$ is a product of a linear factor and an irreducible quartic. If $p\equiv 1$, Frobenius is a translation $i\mapsto i+b$: it is the identity when $b=0$, giving complete splitting, and is a five-cycle when $b\ne 0$, giving an irreducible polynomial.